Prove that $n! > n^5$

Question

I'm trying to prove that $n! > n^5$ for large enough values of n. While it seems obvious that this should be true, I have no idea how to prove it rigorously.

EDIT:

So, looking at the comments, here's what I'm thinking now. What do folks think? Is the following reasonable?

The idea is to take the limit of the ratio of $(n+1)!/n!$ vs $(n+1)^5/n^5$ to show that the former is much larger than the latter, and thus that when n > M, where M is some positive integer, $n!$ must be growing much larger than $n^5$:

Let us assume $\frac{(n+1)!}{n!} > \frac{(n+1)^5}{n^5}$

$$ \lim_{n\to\inf} \frac{(n+1)!}{n!} > \frac{(n+1)^5}{n^5}\\ = \lim_{n\to\inf} n+1 > (\frac{n+1}{n})^5\\ = \lim_{n\to\inf} n+1 > (1 + \frac{1}{n})^5\\ \ldots =\lim_{n\to\inf} n+1 > (1 + 3/n + 8/n^2 + 10/n^3 + 5/n^4 + 1/n^5)\\ = inf > 1 $$

(Edited [again]: I did not know you could use latex)

Answer 1

The first derivative of those should be useful in your proof. Start with n = $10$ for example.

Someone asked me to explain further but I am not a math person so it will be a simple explanation.

For n = $10$ we have $10$! vs. $10^5$ which is $3,628,800$ vs. $100,000$ so it is already larger at the point. If we increase n to $11$, the $10$! increases by a factor of $11$ but the $10^5$ increases by $1.1^5$ which is only about $1.61$. If we increase n to $12$ you can see a "breakaway" between them so just prove this breakaway by showing the "discrete 1st derivative" of n! is increasing greater than that of $n^5$ and you got it.

Answer 2

use induction to prove your inequality for $n\geq 8$

Answer 3

For $n \geq 10$, we have $$n! = n(n-1) \cdots (n-4)(n - 5)! \geq \left(\frac{n}{2}\right)^5 (n - 5)! = n^5\left(\frac{(n -5)!}{32}\right) > n^5.$$ (The bound above is not sharp, but it's sufficient for the problem. Also, the same argument shows that for any $p > 0$, we have $n! > n^p$ for sufficiently large $n$.)

Answer 4

Show $10!>10^5$

Assume $n!>n^5$

Prove $(n+1)!>(n+1)^5$:

• $(n+1)!=n!(n+1)$
• $n!(n+1)>n^5(n+1)$ assumption used here
• $n^5(n+1)=n^6+n^5$
• $n^6+n^5>n^6$
• $n^6>n^5+5n^4+10n^3+10n^2+5n+1$ for large enough values of n
• $n^5+5n^4+10n^3+10n^2+5n+1=(n+1)^5$

Answer 5

Use a multiplicative variant of Gauss's trick: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$ For $n\ge 10$, we get $n! \ge n^5$.

If you really need strict inequality, take $n=11$.

Answer 6

I find the simplest way to prove this sort of inequality is to prove that the ratio of the two sequences converges to $\infty$ or $0$ (depending on which ratio you're looking at). In other words, prove that $\frac {n!} {n^5}\to \infty$.

Answer 7

Think about it this way: $n! = 1 * 2 * 3 * 4 * 5 * ... * n$. We want to prove that it is bigger than $n^5$. That's equivalent to the statement that $n! > (\sqrt n)^{10}$. This means that if we have at least $10$ numbers in $n! = 1 * 2 * 3 * ... * n$ that are strictly greater than $\sqrt n$, the inequality becomes trivial. To illustrate, suppose for simplicity let's set $n=25$. We get $\sqrt n = 5$, so now we have in the product $25-5=20$ multipliers which are at least $\sqrt n = 5$, which is double what we need to prove the inequality.

What we did so far: $n! = 1 * 2 * ... * 5 * 6 * 7 * ... * 14 * 15 * 16 * ... * 25 = 1 * 2 * ... * \sqrt n * (\sqrt n + 1) * (\sqrt n + 2) * ... * (\sqrt n + 9) * (\sqrt n + 10) * 16 * ... * 25 > 1 * 2 * ... * \sqrt n * \sqrt n * \sqrt n * ... * \sqrt n * \sqrt n * 16 * ... * 25 > (\sqrt n)^{10} = n^5$

Generalizing this, we can write: $n! = \prod_{k = 1}^{n}{k} = (\prod_{k = 1}^{\lfloor \sqrt n \rfloor}k)(\prod_{k = \lfloor \sqrt n \rfloor + 1}^{n}k) > \prod_{k = \lfloor \sqrt n \rfloor + 1}^{n}\sqrt n = \sqrt n^{n - \lfloor \sqrt n \rfloor} > \sqrt n^{10} = n^5$

The last inequality is only true when $n-\sqrt n > 10$, but as this goes to infinity when $n \rightarrow \infty$, for some sufficiently large $n$ (and we've seen that $5$ is sufficient), $n! > n^5$. QED.

Answer 8

If $n$ is $10$ or greater, $$\begin{align} n!&=n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)!\\ &=n\big((n-1)(n-5)\big)\big((n-2)(n-6)\big)\big((n-3)(n-7)\big)\big((n-4)(n-8)\big)(n-9)!\\ &>n(n)(n)(n)(n)(n-9)!\\ &=n^5(n-9)!\\ &\geq n^5 \end{align}$$

For $n=9$, $$\frac{9!}{9^5}=\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2}{9^4}=\frac{8\cdot7\cdot2\cdot5\cdot4\cdot2}{9^3}=\frac{560\cdot 8}{729}>1$$ so it is still true.

For $n=8$, $$\frac{8!}{8^5}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2}{8^4}=\frac{7\cdot3\cdot5\cdot3}{8^2\cdot4}=\frac{21\cdot 15}{256}>1$$ so it is still true.

For $n=7$, $$\frac{7!}{7^5}=\frac{6\cdot5\cdot4\cdot3\cdot2}{7^4}=\frac{720}{49^2}<1$$ so it is no longer true.

Therefore $n!>n^5$ for all $n\geq8$.